A line on the wall and a line on the floor are skew. always sometimes never

2 answers:

6 0

They will never intersect nor be parallel.

Is this what you are asking? If they will always, sometimes, or never intersect?

KATRIN_1 [288]2 years ago

3 0

I believe they will sometimes be skew to the floor.

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6x + 9y = -15

tester [92]

Answer:
A
Step-by-step explanation
You can divide 6x,9y and -15 by 3
6x/3=2x
9y/3=3y
-15/3=(-5)
Combine them to get the factored form of 6x+9y-15:
3(2x+3y-5)

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Calculating cos-1 ( help is gladly appreciated :) )

Alekssandra [29.7K]

Answer:

$\frac{3\pi}{4}$

(Assuming you want your answer in radians)

If you want the answer in degrees just multiply your answer in radians by $\frac{180^\circ}{\pi}$ giving you:

$\frac{3\pi}{4} \cdot \frac{180^\circ}{\pi}=\frac{3(180)}{4}=135^{\circ}$ .

We can do this since $\pi \text{ rad }=180^\circ$ (half the circumference of the unit circle is equivalent to 180 degree rotation).

Step-by-step explanation:

$\cos^{-1}(x)$ is going to output an angle measurement in $[0,\pi]$ .

So we are looking to solve the following equation in that interval:

$\cos(x)=-\frac{\sqrt{2}}{2}$ .

This happens in the second quadrant on the given interval.

The solution to the equation is $\frac{3\pi}{4}$ .

So we are saying that $\cos(\frac{3\pi}{4})=\frac{-\sqrt{2}}{2}$ implies $\cos^{-1}(\frac{-\sqrt{2}}{2})=\frac{3\pi}{4}$ since $\frac{3\pi}{4} \in [0,\pi]$ .